Tree Queries - Finding k-th Ancestors

Things to be discussed here,
  • Introduction to Tree Queries
  • Finding Ancestors
  • Explanation with examples
  • Implementation in C++
  • Practice Problems
Introduction to Tree Queries
In the following article, we will be discussing various techniques to process queries on subtrees and paths of a rooted tree. For example, such queries are:
  • What is the kth ancestor of a node?
  • What is the sum of values in the subtree of a node?
  • What is the sum of value on a path between two nodes?
  • what is the lowest common ancestor of two nodes?
So let's start with Finding Ancestors,

Finding Ancestors

The k-th ancestor of a node x in a rooted tree is the node that we will reach if we move k levels up from x. Let ancestor (x,k) denote the kth ancestor of a node x (or 0 if there is no such an ancestor). For example in the following tree ancestor(2,1) = 1 and ancestor(8,2) = 2.

An easy way to calculate any value of ancestor(x,k) is to perform a sequence of k moves in the tree. However, the time complexity of this method is O(k), which may be slow because a tree of n nodes may have a chain of n nodes i.e. linear tree.

Simple Approach to find K-th Ancestor

In this type of problem to find the k-th ancestor in rooted tree

  • First, make an array whose value is the parent of that node.
  • Then for any node first ancestor is its parent, 2nd ancestor is parent of a parent, 3rd ancestor is the parent of parent of parent and so own...
  • For each node, the time taken will be O(logn).

Implementation in C++

For implementation, take input as a tree rooted at node 1

  • In first-line take input n, n - number of nodes in the tree.
  • Next n-1 lines take edges u,v.

Practice Problems

That's all for this article in a later article we will be discussing some different approach, in the next session we will be discussing Sub Tree Queries and problems related to it and for now practice problems. ( If you stuck then do comment Question Number for Solution and your approach so that other programmer or we can help you).

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